The increase in pressure required to decrease the volume of a water sample by 0.2 \% is P \times 10^5 Nm^-2 . Bulk modulus of water is 2.15 \times 10^9 Nm^-2 . The value of P is \underline\hspace2cm .

Properties of Matter PYQ — JEE Main 2025 · 24 Jan · Shift 2 · Q72 with Solution

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Correct answer: 43

The bulk modulus $B$ is defined as

B = -\frac{\Delta P}{\Delta V/V}

For magnitude,

\Delta P = B\left|\frac{\Delta V}{V}\right|

The volume decreases by $0.2\%$ .

0.2\% = \frac{0.2}{100} = 0.002

So,

\left|\frac{\Delta V}{V}\right| = 0.002

Given:

B = 2.15 \times 10^9\ \text{N m}^{-2}

Therefore,

\Delta P = 2.15 \times 10^9 \times 0.002

\Delta P = 2.15 \times 10^9 \times 2 \times 10^{-3}

\Delta P = 4.30 \times 10^6\ \text{N m}^{-2}

It is given that the increase in pressure is

P \times 10^5\ \text{N m}^{-2}

So,

P \times 10^5 = 4.30 \times 10^6

P = \frac{4.30 \times 10^6}{10^5} = 43

\boxed{43}

The derived answer matches the stored correct answer.